Answer:
x=[tex]\frac{16}{\sqrt[3]{2}+1 }[/tex]
Step-by-step explanation:
Q= illumination
I = intensity
Q= I/d^2
Q_total = [tex]\frac{I_1}{d_1^2}+\frac{I_2}{d_2^2}[/tex]
= [tex]\frac{I}{x^2}+\frac{2I}{(16-x)^2}[/tex]
now Q' = 0
⇒I[tex]{-\frac{2}{x^3}}+\frac{4}{(16-x)^3}[/tex]
x=[tex]\frac{16}{\sqrt[3]{2}+1 }[/tex]
[/tex][tex]\frac{1}{x^3} = \frac{2}{(16-x)^3}[/tex]
x=[tex]\frac{16}{\sqrt[3]{2}+1 }[/tex]
is the required point